Math 2250-4
Mon Oct 28
5.3 How to find the solution space for nth order linear homogeneous DE's with constant coefficients, and
why the algorithms work.
, Use Friday's notes to begin discussing the systematic algorithms for finding the general solution yH x
to the constant-coefficient homogeneous linear DE of order n,
L y d y n C an K 1 y n K 1 C... C a1 y#C a0 y = 0 .
Solutions y x to IVPs for these DEs exist and are unique for x-interval I = =, the entire real line.
Finding yH is a key step in finding the general solution y = yP C yH to non-homogeneous DE's L y = f,
and also has interesting applications for just the homogeneous problem, as we will see in section 5.4.
Strategy: In all cases we first try to find a basis for the nKdimensional solution space made of or related to
exponential functions....trying y x = er x yields
L y = er x rn C an K 1 rn K 1 C... C a1 r C a0 = er x p r .
The characteristic polynomial p r and how it factors are the keys to finding the solution space to
L y = 0. There are three cases, of which the first two are on Friday's notes:
Case I p r has n distinct real roots, i.e. p r = r K r1
different from all the others. This is the easiest case.
r K r2 $...$ r K rn with each root rj
Case 2) p r has n real roots, but some of them are repeated roots.
Case 3) p r has complex number roots. This is the hardest, but also most interesting case.
All three cases are important in applications. We should finish the first two cases today, and may have a
chance to start considering Case 3.
Case 3) p r has some complex roots. The punch line is that exponential functions er x still work, except
that r = a G b i ; but, rather than use those complex exponential functions to construct solution space bases
we decompose them into real-valued solutions that are products of exponential and trigonometric
functions.
To understand how this all comes about, we need to learn Euler's formula . This also lets us review some
important Taylor's series facts from Calc 2. As it turns out, complex number arithmetic and complex
exponential functions actually are important in many engineering and science applications.
Recall the Taylor-Maclaurin formula from Calculus
1
1
1
f x w f 0 C f# 0 x C
f## 0 x2 C
f### 0 x3 C....C
f
2!
3!
n!
n
0 xn C....
(Recall that the partial sum polynomial through order n matches f and its first n derivatives at x0 = 0.
When you studied Taylor series in Calculus you sometimes expanded about points other than x0 = 0. You
also needed error estimates to figure out on which intervals the Taylor polynomials actually coverged back
to f .)
Exercise 1) Use the formula above to recall the three very important Taylor series for
1a) ex =
1b) cos x =
1c) sin x =
In Calculus you checked that these series actually converge and equal the given functions, for all real
numbers x.
Exercise 2) Let x = i q and use the Taylor series for ex as the definition of ei q in order to derive Euler's
formula:
ei q = cos q C i sin q .
From Euler's formula it makes sense to define
ea C b i d ea eb i = ea cos b C i sin b
for a, b 2 =. So for x 2 = we also get
e a C b i x = ea x cos b x C i sin b x = ea xcos b x C i ea xsin b x .
For a complex function f x C i g x we define the derivative by
Dx f x C i g x d f# x C i g# x .
It's straightforward to verify (but would take some time to check all of them) that the usual differentiation
rules, i.e. sum rule, product rule, quotient rule, constant multiple rule, all hold for derivatives of complex
functions. The following rule pertains most specifically to our discussion and we should check it:
Exercise 3) Check that Dx e
aCbi x
= aCb i e
aCbi x
, i.e.
Dx er x = r er x
even if r is complex. (So also D2x er x = Dxr er x = r2 er x, D3x er x = r3 er x , etc.)
Now return to our differential equation questions, with
L y d y n C an K 1 y n K 1 C... C a1 y#C a0 y .
Then even for complex r = a C b i (a, b 2 = , our work above shows that
L er x = er x rn C an K 1 rn K 1 C... C a1 r C a0 = er x p r .
So if r = a C b i is a complex root of p r then er x is a complex-valued function solution to L y = 0.
But L is linear, and because of how we take derivatives of complex functions, we can compute in this case
that
0 C 0 i = L er x = L ea xcos b x C i ea xsin b x
= L ea xcos b x C i L ea xsin b x .
Equating the real and imaginary parts in the first expression to those in the final expression (because that's
what it means for complex numbers to be equal) we deduce
0 = L ea xcos b x
0 = L ea xsin b x .
Upshot: If r = a C b i is a complex root of the characteristic polynomial p r then
y1 = ea xcos b x
y2 = ea xsin b x
are two solutions to L y = 0 . (The conjugate root a K b i would give rise to y1 , Ky2 , which have the
same span.
Case 3) Let L have characteristic polynomial
p r = rn C an K 1 rn K 1 C... C a1 r C a0
with real constant coefficients an K 1 ,..., a1 , a0 . If r K a C b i
conjugate factor r K a K b i
to L y = 0, namely
k
k
is a factor of p r
then so is the
. Associated to these two factors are 2 k real and independent solutions
ea xcos b x , ea xsin b x
x ea xcos b x , x ea xsin b x
:
:
ax
ax
xk K 1e cos b x , xk K 1e sin b x
Combining cases 1,2,3, yields a complete algorithm for finding the general solution to L y = 0, as long as
you are able to figure out the factorization of the characteristic polynomial p r .
Exercise 4) Find a basis for the solution space of functions y x that solve
y##C 4 y = 0 .
(You were told a basis in the last problem of last week's hw....now you know where it came from.)
Exercise 5) Find a basis for the solution space of functions y x that solve
y##C 6 y#C 13 y = 0 .
Exercise 6) Suppose a 7th order linear homogeneous DE has characteristic polynomial
2
p r = r2 C 6 r C 13
rK2 3 .
What is the general solution to the corresponding homogeneous DE?